Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $q = \dfrac{z + 7}{z^2 - 12z + 27} \times \dfrac{-5z^2 + 45z}{2z + 14} $
Answer: First factor the quadratic. $q = \dfrac{z + 7}{(z - 9)(z - 3)} \times \dfrac{-5z^2 + 45z}{2z + 14} $ Then factor out any other terms. $q = \dfrac{z + 7}{(z - 9)(z - 3)} \times \dfrac{-5z(z - 9)}{2(z + 7)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ (z + 7) \times -5z(z - 9) } { (z - 9)(z - 3) \times 2(z + 7) } $ $q = \dfrac{ -5z(z + 7)(z - 9)}{ 2(z - 9)(z - 3)(z + 7)} $ Notice that $(z + 7)$ and $(z - 9)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ -5z(z + 7)\cancel{(z - 9)}}{ 2\cancel{(z - 9)}(z - 3)(z + 7)} $ We are dividing by $z - 9$ , so $z - 9 \neq 0$ Therefore, $z \neq 9$ $q = \dfrac{ -5z\cancel{(z + 7)}\cancel{(z - 9)}}{ 2\cancel{(z - 9)}(z - 3)\cancel{(z + 7)}} $ We are dividing by $z + 7$ , so $z + 7 \neq 0$ Therefore, $z \neq -7$ $q = \dfrac{-5z}{2(z - 3)} ; \space z \neq 9 ; \space z \neq -7 $